A Solid-State |
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Placing the ground at the junction of these resistor splits the phase, as the voltage movements oppose themselves when referenced ground, their midpoint. |
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It can't work; its like trying to drive the car forward when its parked in front of a brick wall. Or at least I have been told. But it does work and has been used in a few high-end amplifiers (Acoustac TNT, I believe). The output devices see the same base-to-emitter drive voltages and currents that they would see if the output stage found its ground in the traditional manor, i.e. the center of the two power supplies. Why brother the with this topology? For the same reason we brother with the Circlotron topology, it greatly eases and simplifies the driver portion of the amplifier. In the example above, the driver stage's power supply voltage does not need to equal or exceed the output stage's; in fact, it could be as little as ±10 volts even in a 200 watt amplifier. This is possible because the driver stage only needs to supply a few volts of drive even with a MOSFET based output stage. And lower voltages greatly ease the design of the driver stage. All of which begs the question: if this topology is so good, why isn't it used more often? While the answer has many parts, two play the biggest roles: ignorance and added complexity. Few know about this topology and fewer understand how it works. Second, the circuit requires a separate power supply for the driver stage and, more damningly, a separate power supply per output stage, two channels cannot share the same floating power supply. |
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Making a horizontal split-load phase splitter requires the same three parting of the power supply that the cascode required and the observing of the condition that both load resistors are effectively AC grounded at one end. |
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An additional set of parts has been added by the inclusion of a cathode resistor and its bypass capacitor. These extra components are necessary for a ground level input to match the outputs. If this feature is not necessary, they can be removed. A few readers will question having the inverted output taken indirectly through the power supply. But the secret is that in all circuits shown so far and most other amplifiers, the signal currents travel through the power supplies. |
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